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INT
Group Title
change the cartesian integral into an equivalent polar integral:
 one year ago
 one year ago
INT Group Title
change the cartesian integral into an equivalent polar integral:
 one year ago
 one year ago

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INT Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{\sqrt{2}}^{2}\int\limits_{\sqrt{4y^{2}}}^{y} dx dy\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dw:1353881320734:dwi think this is what it looks like at least
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
remember that x=rcos(t) y=rsin(t) and r=sqrt(x^2+y^2), but thats really already given as 2
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
I think I understand how to get pi/4 < theta < pi/2 But finding the bounds for r is throwing me off.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
well, since r only really changes from 0 to 2; the bounds for r are 0 to 2
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
it r changed according to some function of t the r would be bound from 0 to f(t)
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
according to the answers, r is bounded by 2 < r < 2csc(theta)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dw:1353881730467:dw
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
but at pi/2 r can be 2 and rad(2)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
it might help to see the limits in this fashion to be able to see how to change them out \[\Large \int_{y=\sqrt{2}}^{y=2}~~~\int_{x=\sqrt{4y^{2}}}^{x=y} ~~~dx~dy\]
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
my book says to plug in x=rcos(theta) etc and magically gets the polar bounds. They dont show any intermediate steps so im really confused.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[x= \sqrt{r^2y^2}\]\[2cos(t) = \sqrt{2^2(2sin(t))^2}\]\[4cos^2(t) = 44sin^2(t)\]\[cos^2(t) +sin^2(t)=1~\text{for any t}\] \[x= y\]\[2cos(t) = 2sin(t)\]\[t=\frac{pi}4\] something along those lines i believe
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
y = 2 2sin(t) = 2 sin(t) = 1; when t=pi/2 y=sqrt(2) 2sin(t)=sqrt(2) sin(t)=sqrt(2)/2; when t=pi/4
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i might need to see the steps in your books to refresh me memory
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
this is all that they show.
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
I just dont understand where they cot the csc from.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
your d parts in that png are dydx ... not dxdy
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
well, thats a problem... the question has it as dx dy
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
I dont think dy dx even works. Wouldnt it be y=y y=rad(4y^2)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
yeah, so the png might be a typo
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
ok so ignoring the png, would the r just be 0 to 2?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
from what i can tell yes; dw:1353883154486:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
but then i might have been thinking of this backwards
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
so, r goes from 2 up to the line; its not the area inside the circle
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
if i were to attempt this; I would determine the area of the triangle and subtract out the area inside the circle part
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
But doesnt the outer circle ask that the Y bounds are 2 and rad(2) not the X?
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
oh you flipped it
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i positioned the graph in a convenient manner; a swaped the x and y axis ,,, yes
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
so, what we want is a function in ploar from for the line y=2
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
well, even though the answer key messed up the dxdy, but 2(pi/2) seems to be the right answer.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
it is; and youd be able to see that y=2 converts to 2csc(t) if i could remember how to do it :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
got it r^2 = x^2 + y^2 ; but y=2 is constant r^2 = x^2 + 4 ; x = rcos(t) r^2 = r^2cos^2(t) + 4 r^2r^2cos^2(t) = 4 r^2(1cos^2(t)) = 4 r^2 sin^2(t) = 4 r^2 = 4csc^2(t) r = 2csc(t)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
area of triangle = 2*2/2 = 2 area of sector = 4pi/8 = pi/2 area of shaded region = triangle  sector: 2  pi/2
 one year ago

INT Group TitleBest ResponseYou've already chosen the best response.0
ok that makes sense.. one more question, is it a coincidence that y = 2 r sin(t) = 2 r = 2 / sin(t) using 1/sin(t) = csc(t) trig identity r = 2 csc(t) ?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
not sure if its coincidence or not; but that seems to work out as well :)
 one year ago
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